non-existence of global coordinates

Assume we have a smooth manifold, $M$, of dimension $n$. (An example of interest is the case when $M$ is a compact and orientable Riemann surface of genus $g$, but the question is intended to be broad.)

Then cover $M$ by open sets $\cup_iU_i=M$. In a local coordinate chart, $(U_i,\phi_i)$, where $\phi_i:U_i\rightarrow \mathbb{R}^n$, let us denote these local coordinates by $(\sigma^1,\dots,\sigma^n)$.

My question is: under what conditions do coordinates such as $(\sigma^1,\dots,\sigma^n)$ exist that cover the entire manifold, $M$, and more importantly why?

Related questions are: what is the obstruction to extending the local chart to cover the entire surface (except possibly for a discrete set of points in $M$)? Is there a general reasoning that applies to all cases (at least for the case of an orientable compact Riemann surface)?

Any help much appreciated!

• If you relax "discrete" to "positive codimension", then exponential map is always surjective, if manifold was compact. If, on the contrary, you want existense of single surjective atlas, then I suspect that it's equivalent to manifold being affine (i. e. factor of $\Bbb R^k$ by properly discontinuous action of $Aff(k, \Bbb R)$. – Denis T. Aug 22 '18 at 19:56
• Charts are usually defined to be diffeomorphisms, so you are just asking what are the open subsets of Euclidean space. This of course has nothing to do with any Riemannian metric. What is the role of the Riemannian metric in your question? – Ben McKay Aug 22 '18 at 19:59
• @DenisT.: i'm interested in manifolds with non-trivial Euler class, so doesn't this rules out $Aff(k,\mathbb{R})$? Up to a set of positive co-dimension is ok I think. If we consider the case of a Riemann surface as a complex manifold instead (which we cover with charts with holomorphic transition functions), i read in the literature that there do exist global coordinates modulo $U(1)$ but not global holomorphic coordinates. Any ideas why? thanks! – Wakabaloola Aug 22 '18 at 20:53
• @BenMcKay: Indeed I'm interested in the textbook case where $\phi$ is a diffeomorphism, and the Riemannian metric is indeed redundant for this question (I will update the question to take this into account). thanks! – Wakabaloola Aug 22 '18 at 20:53

• @Wakabaloola: no. Puncturing an $n$-dimensional ball makes a manifold with nontrivial homotopy in dimension $n$, so not diffeomorphic to a ball. A smooth manifold admits global coordinates if and only if it is diffeomorphic to an open subset of Euclidean space. – Ben McKay Aug 23 '18 at 8:29